811Solutions6

# 811Solutions6 - Homework 6 Solutions Kyle Chapman 102.4...

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Homework 6 Solutions Kyle Chapman November 17, 2011 102.4 Seeking a contradiction, suppose there are only ﬁnitely many primes of the form 4 k + 3. Denote them by 4 k 1 + 3 , 4 k 2 + 3 ,..., 4 k n + 3. Multiply all of them together to give N = (4 k 1 + 3)(4 k 2 + 3) ... (4 k n + 3). Expanding this out gives us a number which is odd, so either N = 4 k +1 for some k or N = 4 k +3 for some k . In the ﬁrst case, let M = N + 2 and in the second case let M = N + 4. In both cases, M is of the form 4 j + 3 for some j . We know M = Π l i =1 p i for some collection of primes p i . Further, since M is odd, none of the p i can be 2. This means that every p i is either of the form 4 r + 1 or 4 r + 3. If they are all of the form 4 r + 1 then their product would also be of the form 4 r + 1 which it is not, so at least one of the p i is of the form 4 r + 3. This means that this prime divides both N and M which means it divides their diﬀerence. Their diﬀerence, however, is either 2 or 4 neither of which has a factor of the form 4 r + 3. Thus we have our contradiction. 102.6 From the construction, each of the numbers N,N + 1 ,...,N + 9 is of the form 11! + k where k is between 2 and 11. This means that k ( 11! k

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## This note was uploaded on 12/26/2011 for the course MATH 8 taught by Professor Bigelow during the Fall '08 term at UCSB.

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811Solutions6 - Homework 6 Solutions Kyle Chapman 102.4...

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